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Rizwan Javed could you answer this as well? the last partView attachment 59497
the last part (v), how do I solve this?
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Rizwan Javed could you answer this as well? the last partView attachment 59497
the last part (v), how do I solve this?
z = (X-Mean)/Standard deviation
z value corresponding to X=168 is
z = (168-160)/5 = + 1.6
The area under the standard normal curve right to z = + 1.6 indicates the required probability.
This area lies in the extreme right tail of the normal curve.
The area corresponding to z = 1.6 is 0.4452
The area right to z = 0 is 0.5000
The area beyond z = + 1.6 is or
P(X > 168) = 0.5000 - 0.4452 = 0.0548
Tysm guysμ = 60 g
σ = 5 g
We're to find the probability that X>168, so
P(X>168)
Standardizing X using Z = X- μ / σ
P (Z > 168 -160 / 5)
= P ( Z > 1.6)
= 1- P(Z<1.6)
= 1 - Ф (1.6)
Using Normal Distribution tables, Ф (1.6) = 0.9452
= 1 - 0.9452
= 0.0548 Ans.
Sorry for getting late.Rizwan Javed could you answer this as well? the last part
oh thanks, I think I get that!Sorry for getting late.
First calculate those two expressions given using the standard deviation and mean you calculated in the previous part. Then from the graph find the cumulative frequencies at these two times you have calculated. Then the number of people who entered the store will be:
(cumulative frequency found from the graph at time m +0.5s ) - (lower cumulative frequecy from the graph at time m - 0.5s )
again?View attachment 59503
7(ii) temme ma mistake :
I think it is.Is motion under gravity included in 2016's syllabus for M1?
Hope this helpsView attachment 59503
7(ii) temme ma mistake :
Thank u soooo much
Use time for P to be t+2 and for Q to be t only. You'll get the right answerView attachment 59514
The second part please. I seem to just not get it right! :/
Sp = Sq + 5 ... and then while using s = ut + 0.5at^2 for both sides, we need to use (t - 2) for LHS since Q is projected 2s later ... but doing it that way leads me to a wrong answer.
Here's the ms:
View attachment 59515
View attachment 59517
I had tried that also. But my 5t^2 gets cancelled out and it's just a linear eqn left ... in the ms, there are two values for t.Use time for P to be t+2 and for Q to be t only. You'll get the right answer
Give me 10 minutesI had tried that also. But my 5t^2 gets cancelled out and it's just a linear eqn left ... in the ms, there are two values for t.
Could you please post the solution?
Yeah yeah, no problem!Give me 10 minutes
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